Question

Charge q1 = 29.5 μC is located at r1 = (3.4 i − 3.5 j) m. A second charge q2 = −44.4 μC is located at r2 = (10.5 i + 7.5 j) m. The force of the second charge on the first charge can be written using unit vectors as follows: F = Fx i + Fy j What is the value of Fx in newtons?

Answer 1

Answer:

-0.234 N

Explanation:

Electrostatic force is:

$F=\frac{kq_1q_2}{r^2}$

$k=9\times10^9Nm^2/C^2$

$q_1=29.5\mu C\\q_2=-44.4 \mu C\\r_x^2=(r_{2x}-r_{1x})^2\\=(10.5-3.4)^2=50.41m^2$

Substitute the values:

$F_x=\frac{9\times 10^9\times 29.5\times10^{-6}\times -44.4\times10^{-6}}{50.41}=-0.234N$

Negative sign indicates that the force is attractive in nature which is also true as the charges are opposite in nature.